Olympiad Maths Prep

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Problem 127

AMC 10/12, early questions
Geometry Difficulty 3.5 Prove it National Math Olympiad 2015 – First Round · Slovenia · 2015

We inscribe a regular octagon in a square with side of length aa, so that 4 sides of the octagon lie on the sides of the square. Express the side length of the inscribed octagon in terms of aa.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Denote by xx the side length of the inscribed octagon. The four triangles that are formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length xx, their legs are of length x2\frac{x}{\sqrt{2}}. Thus
a=x+2x2=x+x2=x(2+1). a = x + 2\frac{x}{\sqrt{2}} = x + x\sqrt{2} = x(\sqrt{2} + 1).
From this we deduce

Figure 1
x=a2+1=a(2−1)2−1=2a−a. x = \frac{a}{\sqrt{2} + 1} = \frac{a(\sqrt{2} - 1)}{2 - 1} = \sqrt{2}a - a.

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